Strictly Increasing Function $f$ Satisfying $f(f(x))=3x$ 
Find all strictly increasing functions that satisfy
$$f(f(x))=3x\text.$$

Obviously $\sqrt{3} x$ is an answer, but I am having a hard time proving that is the only solution. Any ideas or solutions? Thanks for the help!
 A: $x\mapsto\sqrt 3x$ is not the only solution. In fact, there are infinitely many. Obviously $f|_{\mathbb R^-}$ and $f|_{\mathbb R^+}$ are independent of each other. So, I'll only consider $f$ on $[0,\infty)$. To construct an alternative solution, let $f : [1,\sqrt 3]\to[\sqrt 3,3]$ be any strictly increasing continuous function with $f(1) = \sqrt 3$ and $f(\sqrt 3) = 3$. Now, for $x\in [\sqrt 3,3]$ set $f(x) := 3f^{-1}(x)$. Then this part of $f$ maps $[\sqrt 3,3]$ to $[3,3\sqrt 3]$. This defines an increasing continuous function from $[1,3]$ to $[\sqrt 3,3\sqrt 3]$.
Now for $x\in [3^k,3^{k+1}]$, $k\in\mathbb Z$, set $f(x) := 3^kf(3^{-k}x)$ and finally $f(0) := 0$. Thus, we have defined a continuous and increasing function on all of $[0,\infty)$. Let us show that it satisfies the functional equation.
To begin with, for $x\in [1,\sqrt 3]$ we have $f(f(x)) = 3f^{-1}(f(x)) = 3x$. Let $x\in [3^k,3^{k}\sqrt 3)$. Then, since $f(3^{-k}x)\in [\sqrt 3,3)$, we have $f(x) = 3^kf(3^{-k}x)\in [3^k\sqrt 3,3^{k+1})$ and so
$$
f(f(x)) = 3^kf(3^{-k}f(x)) = 3^kf(f(3^{-k}x)) = 3x.
$$
Finally, for $x\in [3^k\sqrt 3,3^{k+1})$ (including $k=0$), we have $3^{-k}x\in [\sqrt 3,3)$ and thus $f(3^{-k}x)\in [3,3\sqrt 3)$, i.e., $f(x) = 3^kf(3^{-k}x)\in [3^{k+1},3^{k+1}\sqrt 3)$. Thus,
$$
f(f(x)) = 3^{k+1}f(3^{-k-1}f(x)) = 3^{k+1}f(3^{-k-1}3^kf(3^{-k}x)) = 3^{k+1}f(3^{-1}3f^{-1}(3^{-k}x)) = 3x.
$$
Remark. Instead of $[1,\sqrt 3]$ you can start with any interval $[a,\sqrt 3a]$ and map it increasingly onto $[\sqrt 3a,3a]$. In other words, you can scale everything by $a>0$. Then you have found all possible solutions. And, of course, to make it more general, you can replace $3$ by any positive number.
